Chương 5: ĐẠO HÀM

\(y'=\dfrac{\left(4x+3\right)\left(x^2-5x+2\right)-\left(2x-5\right)\left(2x^2+3x-1\right)}{\left(x^2-5x+2\right)}=\dfrac{-13x^2+10x+1}{\left(x^2-5x+2\right)^2}\)

\(y'=-\dfrac{2x-2}{\left(x^2-2x+5\right)^2}=\dfrac{2-2x}{\left(x^2-2x+5\right)^2}\)

\(y'\ge0\Leftrightarrow\dfrac{2-2x}{\left(x^2-2x+5\right)^2}\ge0\Rightarrow x\le1\)

Có \(1-\left(-8\right)+1=10\) số nguyên

\(f'\left(x\right)=\dfrac{3x^2\left(x-1\right)-1.x^3}{\left(x-1\right)^2}=\dfrac{2x^3-3x^2}{\left(x-1\right)^2}\)

\(f'\left(x\right)=3x^2-6x\)

\(f'\left(x\right)< 0\Leftrightarrow3x^2-6x< 0\Rightarrow0< x< 2\)

\(f'\left(x\right)=2m-3mx^2\Rightarrow f'\left(1\right)=2m-3m=-m\)

\(\Rightarrow-m\le1\Rightarrow m\ge-1\)

Hàm \(y=\dfrac{ax+b}{cx+d}\) có đạo hàm \(y'=\dfrac{ad-bc}{\left(cd+d\right)^2}\)

Do đó: \(y'=\dfrac{5m-n}{\left(x+5\right)^2}\)

\(y=10\) là hàm hằng nên \(y'=0\) với mọi x

\(f'\left(x\right)=x^2-4\sqrt{2}x+8=\left(x-2\sqrt{2}\right)^2\)

\(f'\left(x\right)=0\Rightarrow\left(x-2\sqrt{2}\right)^2=0\Rightarrow x=2\sqrt{2}\)

\(f'\left(x\right)=-\dfrac{1}{x^2}\Rightarrow f'\left(\sqrt{2}\right)=-\dfrac{1}{\left(\sqrt{2}\right)^2}=-\dfrac{1}{2}\)

\(y=x^2+5+\dfrac{m}{x}\Rightarrow y'=2x-\dfrac{m}{x^2}\)

\(y'=2x+\dfrac{1}{x^2}\Leftrightarrow m=-1\)